south axholme school as paper 1 and 2 energetics … axholme school page 1 as paper 1 and 2 –...

South Axholme School

AS Paper 1 and 2 – Energetics MARK SCHEME

M1.D [1]

M2.D [1]

M3.A [1]

M4.D [1]

M5.B [1]

M6.C [1]

M7. (a) (Energy required) to break a given covalent bond (1) averaged over a range of compounds (1)

Penalise first mark if ‘energy’ / ‘enthalpy’ evolved 2

(b) (i) 4 × C−H = 4 × 413 = +1652 1 × C−C = 1 × 347 = 347 1 × C=O = 1 × 736 = 736 2½ × O=O = 2.5 × 498 = 1245 (1) = 2735 + 1245 = +3980 (1)

first mark for 4 : 1: 1 or 2735 ignore sign


(ii) 4 × H−O = -4 × 464 = –1856 4 × C−O = -4 × 736 = –2944 (1) = –4800 (1)

First mark for 4 : 4

(iii) ΔHR = ΣBonds broken − ΣBonds made = +3980 − 4800 = −820 (1)

Conseq Mark for incorrect answers in (i) and (ii) as (i) Answer + (ii) Answer =

5

[7]

M8. (a) (Enthalpy change) when 1 mol (1) of a compound is formed from its constituent elements (1) in their standard states (1)

3

Allow energy or heat, Ignore evolved or absorbed Mark each point independently

(b) (The enthalpy change for a reaction is) independent of the route (1) 1

(c) ΔHR = Hf products ‑ Hf reactants (1)

= [(3 × -286) + (3 × -394)] ‑ (-248) (1)

= -1792 (1) (kJ mol–1)

Deduct one mark for each error to zero 3

[7]

M9.(a) 2AgNO3 + Zn → Zn(NO3)2 + 2Ag (1)

Accept an ionic equation i.e.2Ag+ +Zn → 2Ag + Zn2+


1

(b) Moles = mv / 1000 (1) = 0.20 × 50/1000 = 1.00 × 10–2

2

(c) Heat energy change = mCΔT (1) = 50 × 418 × 3.2 J

= 669 J (Ignore signs) (1)

Allow 668, 67.0 0.67kJ

Penalise wrong units if given 2

(d) = 134 kJ mol–1

Mark one : 2 × (answer to (c))

Mark two : Dividing by answers to (b)

Allow 133 – 134

Penalise incorrect units

Mark conseq to equation in (a) for full marks, also to that in (c)

If No working is shown and answer is incorrect zero 2

(e) Incomplete reaction or Heat loss (1) 1

[8]

M10. (a) Heat energy change (1)

Not energy on its own

measured at constant pressure (1)

Mark separately, ignore constant temperature statements 2

(b) (i) Enthalpy change when 1 mol of a substance (or compound / product) (1) is formed from its constituent elements (1) in their standard states (1) under standard conditions (1)

Mark separately


(ii) 2Na(s) + S(s) + 2O2(g) → Na2SO4(s)

Balanced (1) State symbols (1), but only if all species are correct

Allow S8 (s) 5

(c) Enthalpy change is independent of reaction route (1)

Penalise incorrect additional statements 1

(d)

–1356 + (2 × 285.8) + (4 × 393.5) + ΔHfC4H4O4 = 0

ΔHf = –789.6 kJ mol–1

If answer is incorrect:

Score +789.6 two marks

Score (× 1); (× 2) and (× 4) for species - one mark

If an incorrect negative answer given check for AE for loss of one mark

3

[11]

M11.(a) Enthalpy (Energy) to break a (covalent) bond (1) OR dissociation energy Varies between compounds so average value used (1) QL mark

OR average of dissociation energies in a single molecule / e.g. CH4

Do not allow mention of energy to form bonds


but with this case can allow second mark otherwise 2nd mark consequential on first

2

(b) (i) 1/2 N2 + 3/2 H2 → NH3 (1)

Ignore s s

(ii) ΔH = (Σ)bonds broken – (Σ)bonds formed (1) = 1/2 × 944 + 3/2 × 436 – 3 × 388 (1) = –38 kJ mol–1 (1)

Ignore no units, penalise wrong units Score 2/3 for -76 1/3 for +38 Allow 1/3 for +76

4

(c) 4 (C–H) + (C=C) + (H–H) – (6 (C–H) + (C–C)) = –136 (1) OR (C=C) + (H–H) – ((C–C) + 2 (C–H)) = –136 2 (C–H) = 836 (1) (C–H) = 418 (kJ mol–1) (1)

Note: allow (1) for –836 another (1) for –418

3

[9]

M12. (a) They are elements (1)

Ignore irrelevant comments 1

(b) Enthalpy change (1)

or heat energy change or heat change or ΔH or any named enthalpy change C.E. if change not mentioned

Independent of route (1)

OR depends on initial and final states Only give second mark if first mark awarded except allow if energy used instead of enthalpy

2

(c) ΔH = ΣΔHf (products) ‑ ΣΔHf (reactants) (1) (Or a cycle)

= 2 × –242 + ½ × –394 – (–365) (1) (also implies first mark)


= -316 kJ mol–1 (1) 3

Ignore no units penalise wrong units +316 scores 1/3

[6]

M13.C

[1]

M14. (a) C3H6O + 4O2 → 3CO2 + 3H2O (1) (or multiple) 1

(b) (i) (1) = 0.0250 (1)

allow 0.025 allow conseq on wrong Mr

1.45/100, CE; C.E.

(ii) heat released = mcΔT = 100 × 4.18 × 58.1 (1)

if 1.45 used in place of 100 CE = 0

= 24300 J (1) (or 24.3kJ)

allow 24200 to 24300 ignore decimal places units tied to answer

If use 0.1 × 4.18 × 51.8 allow ½ for 24.3 with no units

(iii) = –972 (kJ mol–1) (1)

allow –968 to –973 allow +972 allow conseq allow no units


penalise wrong units 5

(c) (i) Heat loss (1) or energy loss

do not allow incomplete combustion

(ii) Difference: more negative (1) (or more exothermic)

QoL mark

Explanation: heat (or energy) released when water vapour condenses (1) or heat/energy required to vaporise water or water molecules have more energy in the gaseous state

3

(d) ΔH = ΣΔHreactants – ΣΔHproducts (1)

(or cycle ) = (2 × –394) + (3 × –286) + (–297) – (–1170) (1) = –773 (1)

ignore units even if wrong

Allow 1/3 for +773 3

[12]

M15.A

[1]

M16. (a) Standard enthalpy change, ΔH : ΔHR = ΣΔHfproducts - ΣΔHfreactants (1)

or cycle

ΔHR = (0 + [2 × –242]) – (4 × –92) (1) = -484 + 368 = –116 (kJ mol–1)

Allow max 1 for +116


Standard entropy change, ΔS : ΔS = ΣΔHf products – ΣΔHf reactants

ΔS = ([2 × 223] + [2 × 189]) - (205 + [4 × 187]) (1)

= 824 – 953

= –129 (J K–1 mol–1)

allow max one for +129 6

(b) (i) Effect: Equilibrium displaced to right / to products (1) Explanation: Reaction is endothermic (1) Constraint reduced (1)

mark separately

(ii) Feasible when ΔG 0 (1)

ΔG = ΔH – TΔS (1)

T = ΔH/ΔS = 208 × 1000 (1) / 253

= 822 K (1) 7

[13]

M17. (a) (i) enthalpy (or heat or heat energy) change when 1 mol of a substance (1) (QL mark) is formed from its elements (1) all substances in their standard states (1) (or normal states at 298K, 100 kPa or std condits)

not STP, NTP 3

(b) enthalpy change (or enthalpy of reaction) is independent of route (1)

ΔH = ΣΔHf prods - ΣΔHf reactants (or cycle) (1) minimum correct cycle is:

ΔH = -642 – 286 – (–602 + 2 × –92) (1) = –142 (kJ mol–1) (1)

penalise this mark for wrong units


+142 scores 1 mark out of the last three 4

(c) ΔH = mcT (1) (or mcΔT) = 50 × 4.2 × 32 = 6720 J = 6.72J (1)

mark is for 6720 J or 6.72 kJ

moles HCl = × conc = × 3 (1)

= 0.15 (1)

if error here mark on conseq.

Therefore moles of MgO reacted = moles HCl/2 (1) (mark is for/2, CE if not/2) = 0.15/2 = 0.075

Therefore ΔH = 6.72/0.075 (1) = –90 kJ (mol–1)

kJ must be given, allow 89 to 91

value (1) sign (1); this mark can be given despite CE for /2

8

Note various combinations of answers to part (c) score as follows:

–89 to –91 kJ (8) (or –89000 to 91000J)

no units (7)

+89 to +91 kJ (7) (or + 89000 to +91000J)

no units (6)

–44 to –46 kJ (5) (or -44000 to -46000J)

no units (4) if units after 6.72 or 6720 (5)

+44 to +46 kJ (4) (or +44000 to + 46000)

if no units and if no units after 6.72 or 6720 (3) otherwise check, could be (4)

[15]

M18.C

[1]


M19.D [1]

M20. (a) M1 Kp = (PY)3. (PZ)2/ (PW)2.( PX) NB [ ] wrong 1

M2 temperature 1

M3 increase 1

M4 particles have more energy or greater velocity/speed 1

M5 more collisions with E > Ea or more successful collisions 1

M6 Reaction exothermic or converse 1

M7 Equilibrium moves in the left 1

Marks for other answers Increase in pressure or concentration allow M1, M5, M6 Max 3 Addition of a catalyst; allow M1, M5, M6 Max 3 Decrease in temperature; allow M1, M2, M6 Max 3 Two or more changes made; allow M1, M6 Max 2

(b) (i) Advantage; reaction goes to completion, not reversible or faster

1

Disadvantage; reaction vigorous/dangerous

(exothermic must be qualified)

or HCl(g) evolved/toxic or CH3COCl expensive

NB Allow converse answers Do not allow reactions with other reagents e.g. water or ease of separation

1

(ii) ΔS = ΣS products – ΣS reactants 1


ΔS = (259 + 187) – (201 + 161) 1

ΔS = 84 (JK–1 mol–1) (Ignore units)

Allow – 84 to score (1) mark 1

ΔG = ΔH – TΔS 1

= – 21.6 – 298 × 84/1000 = – 46.6 kJ mol–1 or – 46 600 J mol–1

1

Allow (2) for – 46.6 without units

(Mark ΔG consequentially to incorrect ΔS)

(e.g. ΔS = –84 gives ΔG = +3.4 kJ mol–1) 1

[15]

M21. (a) (i) ΔH atomisation/sublimation of magnesium 1

(ii) Bond/dissociation enthalpy of Cl-Cl

OR 2 × H atomisation of chlorine 1

(iii) Second ionisation enthalpy of magnesium 1

(iv) 2 × electron affinity of chlorine 1

(v) Lattice formation enthalpy of MgCl2

1

(b) Equation 2MgCl(s) → MgCl2(s) + Mg(s)

State symbols not required but penalise if incorrect 1

Calculation ΔH reaction = ΣΔHf products – ΣΔHf reactants 1

= – 653 – (2 × –133) 1


= – 427 (kJmol–1)

Allow +427 to score (1) mark

Other answers; award (1) for a correct H reaction expression 1

(c) ΔH soln MgCl2 = – ΔH Lat.form. + ΔH hyd.Mg2+ + 2ΔH hyd.Cl–

1

or cycle = 2502 – 1920 – (2 × 364)

1

= – 146 (kJmol–1)

Allow + 146 to score (1) mark

Other answers; award (1) for a correct ΔH soln MgCl2 expression/cycle

1

[12]

M22. (a) ΔH = Σ(bonds broken) – Σ(bonds formed) (or cycle) 1

= +146 – 496/2 (or 2 × 463 + 146 –(2 × 463 + 496/2) 1

= – 102 (kJ mol–1) (1)

(accept no units, wrong units loses a mark; +102 scores (1) only) 1

(b) C(s) + 2H2(g) → CH4(g) equation (1) Correct state symbols (1) 2

(c) (i) Macromolecular

(accept giant molecule or carbon has many (4) bonds) 1

(ii) ΔH = ΣΔHf(products) – ΣΔHf (reactants) (or cycle) 1

= 715 + 4 × 218 – (–74.9) 1

= 1662 (kJ mol–1)


(accept no units, wrong units loses one mark, allow 1660 to 1663, –1662 scores one mark only)

1

(iii) 1662/4 = 415.5

(mark is for divide by four, allow if answer to (c)(ii) is wrong) 1

[10]

M23.A

[1]

M24.B [1]

M25. (a) (i) enthalpy change when 1 mol of a substance (or compound) (QL mark)

1

is (completely) burned in oxygen (or reacted in excess oxygen) 1

at 298 K and 100 kPa (or under standard conditions) 1

(ii) heat produced = mass of water × Sp heat capacity xΔT (or mcΔT)

1

= 150 × 4.18 × 64 (note if mass = 2.12 lose first 2 marks then conseq) = 40100 J or = 40.1 kJ (allow 39.9 - 40.2 must have correct units)

1

moles methanol = mass/Mr = 2.12/32 (1) = 0.0663

1

ΔH = – 40.1/0.0663 = – 605 kJ (mol–1) 1

(allow –602 to –608 or answer in J)

(note allow conseq marking after all mistakes but


note use of 2.12 g loses 2 marks

(b) (i) equilibrium shifts to left at high pressure 1

because position of equilibrium moves to favour fewer moles (of gas)

1

(ii) at high temperature reaction yield is low (or at low T yield is high) 1

at low temperature reaction is slow (or at high T reaction is fast) 1

therefore use a balance (or compromise) between rate and yield 1

(c) ΔH = ΣΔHcο(reactants) – ΣΔHc

ο (products) (or correct cycle) 1

ΔHcο (CH3OH) = ΔHc

ο(CO) + 2 × ΔHcο(H2) – ΔH

1

= (–283) + (2 × –286) – (–91) (mark for previous equation or this) = –764 (kJ mol–1) ( units not essential but lose mark if units wrong) (note + 764 scores 1/3)

1

[15]

M26.C

[1]

M27. (a) enthalpy (or energy) to break (or dissociate) a bond; 1

averaged over different molecules (environments); 1

enthalpy (or heat energy) change when one mole of a compound; 1

is formed from its elements; 1


in their standard states; 1

(b) enthalpy change = Σ(bonds broken) – Σ(bonds formed) or cycle; 1

= 4 × 388 +163 + 2 × 146 + 4 × 463 – (944 + 8 × 463); (or similar)

1

= –789;

(+ 789 scores 1 only) 1

(c) (i) zero; 1

(ii) AH = Σ (enthalpies of formation of products) –Σ (enthalpies of formation of reactants)

1

= 4 × –242-(75 + 2 × –133); 1

= –777;

(+ 777 scores one only) 1

(d) mean bond enthalpies are not exact

(or indication that actual values are different from real values) 1

[13]

M28. (a) enthalpy change/ heat energy change when 1 mol of a substance 1

is completely burned in oxygen 1

at 298K and 100 kPa or standard conditions 1

(not 1atm)


(b) ∆H = ∑ bonds broken – ∑ bonds formed 1

= (6 × 412) + 612 +348 + (4.5 × 496) – ((6 × 743) + (6 × 463)) 1

= – 1572 kJ mol–1

1

(c) by definition ∆Hf is formation from an element 1

(d) ∆Hc = ∑ ∆Hf products -∑ ∆Hf reactants or cycle 1

= (3 × – 394) + (3 × –242) – (+20) 1

= − 1928 kJ mol–1

1

(e) bond enthalpies are mean/average values 1

from a range of compounds 1

[12]

M29.A

[1]

M30. (a) Particles are in maximum state of order

(or perfect order or completely ordered or perfect crystal or minimum disorder or no disorder)

(entropy is zero at 0 k by definition) 1

(b) (Ice) melts

(or freezes or changes from solid to liquid or from liquid to solid) 1


(c) Increase in disorder 1

Bigger (at T2) 1

Second mark only given if first mark has been awarded

(d) (i) Moles of water = 1.53/18 (= 0.085) 1

Heat change per mole = 3.49/0.085 = 41.1 (kJ mol–1)

(allow 41 to 41.1, two sig. figs.)

(penalise –41 (negative value), also penalise wrong units but allow kJ only)

1

(ii) ΔG = ΔH – TΔS 1

(iii) ΔH = TΔS or ΔS = ΔH/T

(penalise if contradiction) 1

ΔS = 41.1/373 = 0.110 kJ K–1 (mol–1) (or 110 (J K–1 (mol–1))

(allow 2 sig. figs.)

(if use value given of 45, answer is 0.12 (or 120 to 121)

(if ΔH is negative in (d) (i), allow negative answer)

(if ΔH is negative in (d) (i), allow positive answer)

(if ΔH is positive in (d) (i), penalise negative answer) 1

Correct units as above (mol–1 not essential) 1

[10]

M31.(a) Enthalpy change when 1 mol of compound (1)

Is formed from it’s elements (1)

All substances in their standard state (1) 3

(b) ΔH = ΣΔHοc (reactants) – ΣΔHο

c (products) (1)


= (7x – 394) + (4 x – 286) – (– 3909) (1)

= + 7 kJmol–1 (1) 3

(c) Heat change = m c ΔT (1)

= 250 × 4.18 × 60 = 62700J = 62.7kJ (1)

Moles C7H8 = 2.5 /92 = 0.0272 (1)

ΔH = 62.7 / 0.0272 = – 2307 kJ mol–1 (1)

(allow –2300 to –2323) 4

(d) Mass of water heated = 25 + 50 = 75g Temp rise = 26.5 – 18 = 8.5 °C

both for (1) mark

Heat change = 75 × 4.18 × 8.5 = 2665 J = 2.665 kJ (1)

Moles HCl = 0.05 (1)

ΔH = – 2.665 / 0.05 = –53.3 kJmol–1 (1)

(allow –53 to –54) 4

(e) Less heat loss (1) 1

[15]

M32. (a) Equation 1/2N2 + 3/2H2 → NH3

1

ΔHf = [(945 × 0.5) + (426 × 1.5)] – (391 × 3) 1

= –46.5 kJ mol–1

1


Mark Range

The marking scheme for this part of the question includes an overall assessment for the Quality of Written Communication (QWC). There are no discrete marks for the assessment of QWC but the candidates’ QWC in this answer will be one of the criteria used to assign a level and award the marks for this part of the question

Descriptor an answer will be expected to meet most of the criteria in the level

descriptor

4-5 – claims supported by an appropriate range of evidence

– good use of information or ideas about chemistry, going beyond those given in the question

– argument well structured with minimal repetition or irrelevant points

– accurate and clear expression of ideas with only minor errors of grammar, punctuation and spelling

2-3 – claims partially supported by evidence

– good use of information or ideas about chemistry given in the question but limited beyond this

– the argument shows some attempt at structure

– the ideas are expressed with reasonable clarity but with a few errors of grammar, punctuation and spelling

0-1 – valid points but not clearly linked to an argument structure

– limited use of information or ideas about chemistry

– unstructured

– errors in spelling, punctuation and grammar or lack of fluency

(b) The higher the temperature the faster the reaction QWC 1

but, since the reaction is exothermic 1

the equilibrium yield is lower QWC 1

The higher the pressure the greater the equilibrium yield QWC 1

because there is a reduction in the number of moles of gas in the reaction

1

but higher pressure is expensive to produce or plant is more


expensive to build QWC 1

A better catalyst would lessen the time to reach equilibrium 1

and allow more ammonia to be produced in a given time QWC 1

[11]

M33. (a) The enthalpy change when 1 mol of a compound 1

is completely burnt in oxygen 1

under standard conditions, or 298K and 100kPA 1

(b) (i) C2H6 + 3½O2 → 2CO2 + 3H2O 1

(ii) ΔH = 2 × ΔHfο (CO2) + 3 × ΔHf

ο (H2O) – ΔHfο (C2H6)

1

= – 788 – 858 – (–85) 1

= – 1561 kJ mol–1

1

(c) moles methane = = 6.25 × 10–3

1

kJ evolved = 6.25 × 10–3 × 890 = 5.56 1

5.56 × 103 joules = (mc)ΔT 1

ΔT = = 46.4 K 1

[11]


M34. (a) (i) q = mc ΔT

Ignore case except T 1

(ii) 8.80 × 1.92 × 9.5 = 161 (J) to 160.5(12) (J)

Credit 0.161 provided it is clear that it is kJ. Penalise wrong units

1

(iii) 11.95 × 0.96 × 9.5 = 109 (J) to 108.98(4) (J)

Credit 0.109 provided it is clear that it is kJ. Penalise wrong units.

1

(iv) M1 Addition of (a)(ii) and (a)(iii)

M2 Multiply by 10 and convert to kJ (divide by 1000) leading to an answer

Consequential on (a)(ii) and (a)(iii) Penalise wrong units Ignore the sign

Therefore ΔH = (–) 2.69 OR (–) 2.7(0) (kJ mol–1)

Ignore greater numbers of significant figures (2.69496) Subtraction in M1 is CE

2

(b) One from:

• No account has been taken of the intermolecular forces initially in the two liquids OR each liquid has its own intermolecular forces in operation before mixing.

• The liquids may react or reference to reaction or reference to bonds broken or formed

Any statement which shows that there are other intermolecular forces to consider. Ignore heat loss and ignore poor mixing.

1

[6]


M35. (a) (i) q = mc ΔT

Ignore case except T 1

(ii) 8.80 × 1.92 × 9.5 = 161 (J) to 160.5(12) (J)

Credit 0.161 provided it is clear that it is kJ. Penalise wrong units

1

(iii) 11.95 × 0.96 × 9.5 = 109 (J) to 108.98(4) (J)

Credit 0.109 provided it is clear that it is kJ. Penalise wrong units.

1

(iv) M1 Addition of (a)(ii) and (a)(iii)

M2 Multiply by 10 and convert to kJ (divide by 1000) leading to an answer

Consequential on (a)(ii) and (a)(iii) Penalise wrong units Ignore the sign

Therefore ΔH = (–) 2.69 OR (–) 2.7(0) (kJ mol–1)

Ignore greater numbers of significant figures (2.69496) Subtraction in M1 is CE

2

(b) One from:

• No account has been taken of the intermolecular forces initially in the two liquids OR each liquid has its own intermolecular forces in operation before mixing.

• The liquids may react or reference to reaction or reference to bonds broken or formed

Any statement which shows that there are other intermolecular forces to consider. Ignore heat loss and ignore poor mixing.

1

[6]

south axholme school as paper 1 and 2 energetics … axholme school page 1 as paper 1 and 2 –...

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